Phương trình:${\sin ^{12}}x + {\cos ^{12}}x = 2({\sin ^{14}}x + {\cos ^{14}}x) + \frac{3}{2}{\rm{cos}}2x$ có nghiệm là
A. $x = \frac{\pi }{4} + k\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{\pi }{4} + k\frac{\pi }{2}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{4} + k2\pi $, $k \in \mathbb{Z}$.
D. Vô nghiệm.
A. $x = \frac{\pi }{4} + k\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{\pi }{4} + k\frac{\pi }{2}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{4} + k2\pi $, $k \in \mathbb{Z}$.
D. Vô nghiệm.
Chọn B
${\sin ^{12}}x + {\cos ^{12}}x = 2({\sin ^{14}}x + {\cos ^{14}}x) + \frac{3}{2}{\rm{cos}}2x$
$ \Leftrightarrow {\sin ^{12}}x\left( {1 - 2{{\sin }^2}x} \right) + {\cos ^{12}}x\left( {1 - 2{{\cos }^3}x} \right) = \frac{3}{2}{\rm{cos}}2x$
$ \Leftrightarrow {\sin ^{12}}x.\cos 2x - {\cos ^{12}}x.\cos 2x = \frac{3}{2}{\rm{cos}}2x$$ \Leftrightarrow \cos 2x\left( {{{\sin }^{12}}x - {{\cos }^{12}}x - \frac{3}{2}} \right) = 0$
$ \Leftrightarrow \cos 2x = 0$ vì ${\sin ^{12}}x - {\cos ^{12}}x \le {\sin ^2}x + \cos {x^2} = 1 < \frac{3}{2}$$ \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\,\,\,(k \in \mathbb{Z})$
${\sin ^{12}}x + {\cos ^{12}}x = 2({\sin ^{14}}x + {\cos ^{14}}x) + \frac{3}{2}{\rm{cos}}2x$
$ \Leftrightarrow {\sin ^{12}}x\left( {1 - 2{{\sin }^2}x} \right) + {\cos ^{12}}x\left( {1 - 2{{\cos }^3}x} \right) = \frac{3}{2}{\rm{cos}}2x$
$ \Leftrightarrow {\sin ^{12}}x.\cos 2x - {\cos ^{12}}x.\cos 2x = \frac{3}{2}{\rm{cos}}2x$$ \Leftrightarrow \cos 2x\left( {{{\sin }^{12}}x - {{\cos }^{12}}x - \frac{3}{2}} \right) = 0$
$ \Leftrightarrow \cos 2x = 0$ vì ${\sin ^{12}}x - {\cos ^{12}}x \le {\sin ^2}x + \cos {x^2} = 1 < \frac{3}{2}$$ \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\,\,\,(k \in \mathbb{Z})$